130_notes.dvi

Together, these areexactly equivalent to the set of solutions we had with the boxdefined
to be from 0 toa. Theu+n(x) have eigenvalue +1 under the parity operator. Theu−n(x) have
eigenvalue -1 under the parity operator.

This is anexample of a symmetryof the problem, causing an operator to commute with the
Hamiltonian. We can then have simultaneous eigenfunctions of that operator andH. In this case
all the energy eigenfunctions are also eigenstates of parity. Parity is conserved.

An arbitrary wave function can be written as a sum of the energy eigenfunctions recovering the
Fourier series in its standard form.

ψ(x) =

∑∞

n=1

[A+nu+n(x) +A−nu−n(x)]

8.6 Momentum Eigenfunctions

We can also look at theeigenfunctions of the momentum operator.

popup(x) =pup(x)

̄h

i

d

dx

up(x) =pup(x)

Theeigenstatesare
up(x) =Ceipx/ ̄h

withpallowed to be positive or negative.

These solutions do not go to zero at infinity so they are not normalizable to one particle.

〈p|p〉=〈up|up〉=∞

This is a common problem for this type of state.

We will use adifferent type of normalization for the momentum eigenstates(and the
position eigenstates).

〈p′|p〉=|C|^2

∫∞

−∞

ei(p−p

′)x/ ̄h

dx= 2π ̄h|C|^2 δ(p−p′)

Instead of the Kronecker delta, we use the Dirac delta function. The momentum eigenstates have
a continuous range of eigenvalues so that they cannot be indexed like the energy eigenstates of a
bound system. This means the Kronecker delta could not work anyway.

These are themomentum eigenstates

up(x) =

1

√

2 π ̄h

eipx/ ̄h